有没有“顶介子”？
近来学习粒子物理，注意到介子的结构很有意思，如：(加'表示反粒子)
轻介子： du'﹑d'u
奇异介子：ds'﹑d's﹑us'﹑u's
粲介子： dc'﹑d'c﹑uc'﹑u'c﹑sc'﹑s'c
底介子： db'﹑d'b﹑ub'﹑u'b﹑sb'﹑s'b
前两种介子由夸克d、u、s及其反粒子组成，后两种介子由夸克d、u、s及其反粒子加粲夸克c和底夸克b及其反粒子组成。照这样下去，应该还存在一种“顶介子”，其组成为：
顶介子：dt'﹑d't﹑ut'﹑u't﹑st'﹑s't（t表示顶夸克）
若有，则顶介子的主要性质应为：质量20000MeV左右，寿命3*10^-19秒左右，自旋0，宇称-,主要衰变物为底介子B和轻介子π。
各位老师和仁兄，有这种可能性吗？
仅为砖瓦之论，愿听珠玉之言！
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一叶孤舟  发表文章数: 38
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 忍一时风平浪静 退一步海阔天空
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sage 发表文章数: 359
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若有，则顶介子的主要性质应为：质量20000MeV左右，寿命3*10^-19秒左右，自旋0，宇称-,主要衰变物为底介子B和轻介子π。
======================================================================= The time scale it takes for a quark to form a meson is about the inverse of the QCD scale 1/(several hundred MeV). Top quark lifetime, on the other hand, is about 1/(several GeV). Therefore, top quark typically decays before it forms any bound states with other quarks. There is no top-meson. By the way, since top quark mass is about 175 GeV, even if it forms a meson, its mass should be about, in your unit, 170000-180000 MeV.
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一叶孤舟 发表文章数: 38
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谢谢sage老师！您的指导使我受益非浅，只是还有一些疑问需要请教。您说顶夸克的质量大约175GeV，若顶夸克构成介子，则这个介子的质量将达170000-180000MeV。可是，QCD上有一个结论，说夸克的质量是渐近于无的。这是否可以理解为：若两个夸克相距非常近，则它们构成的介子的质量就不一定太大。另外，介子的寿命与构成它的夸克的寿命之间的联系是否是必然的？
忍一时风平浪静 退一步海阔天空
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sage 发表文章数: 359
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可是，QCD上有一个结论，说夸克的质量是渐近于无的。
there is no such a conclusion in QCD. QCD says that its coupling constant become small when the quarks get close. 这是否可以理解为：若两个夸克相距非常近，则它们构成的介子的质量就不一定太大。 另外，介子的寿命与构成它的夸克的寿命之间的联系是否是必然的？ Yes. If the quark decays, the meson will become something else. It does not matter whether it is in a bound state or not.
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一叶孤舟 发表文章数: 38
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Sage老师，您说的“渐近自由”，在QCD上用公式表示是
limg＾2(Q＾2)=0（当Q＾2→∞时）（Q表示动量，g表示耦合常数） 但是，在QCD上的确还有一个公式，就是 lim m(Q)=0 (当Q＾2→∞时) 前一个公式表明：当动量很大，距离很近时，夸克、胶子间的相互作用趋于零。 后一个公式表明：当动量很大，距离很近时，夸克的质量趋于零。同时还特别指出，这是胶子自作用超过它和夸克作用的效应，不是相对论效应。 我的理解是：作用的“渐近自由”与质量的“渐近于无”之间有一种密切的联系。 另外，QCD在解释强作用的所谓硬过程方面，取得了很大的成功，但在阐述软过程方面进展缓慢，在夸克囚禁问题上还缺少解决办法。因此，它对粒子质量的描述远不如GWS精确。 一旦“进入”粒子“内部”，各种不确定性就会越来越大，而各种可能性就会越来越多，所谓的“虚过程”将越来越重要，依据不确定关系，在Δt时间内允许存在不守恒量ΔΕ。汤川因此得到了第一个介子――π介子，后来又发现了K介子、D介子和B介子，为什么自然界要有这么多介子哪？我的理解是：因为有六种（味）夸克，所以至少要有五种介子。 肤浅之论，诚心求教。 忍一时风平浪静 退一步海阔天空
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sage 发表文章数: 359
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>lim m(Q)=0 (当Q＾2→∞时)
where do you see this formula? In what context?
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一叶孤舟 发表文章数: 38
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在华东师大出版社1984年出版的胡瑶光的＜规范场论＞第256页，在“奇迹文库”有网络版。这个版本早一些，可能现在有新理论，我还得好好学学。
忍一时风平浪静 退一步海阔天空
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sage 发表文章数: 359
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There is no new theory of QCD. However, the statement you have quoted is either wrong, or in some special context referring to something else. That's why I asked you what is the CONTEXT in which you have found this statement.
Anyway, it is not that important to your question. There is no top meson because 1) top quark decays much before it could have time to form any meson. 2) Even if top-quark will form some bound state, the Q relevant for that is about m_t=175 GeV. QCD coulping at this scale is weak, incapable of forming bound state. 在华东师大出版社1984年出版的胡瑶光的＜规范场论＞第256页，在“奇迹文库”有网络版。这个版本早一些，可能现在有新理论，我还得好好学学。 By the way, this is one of the worst field theory books I have ever seen.
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卢昌海  发表文章数: 768
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一叶孤舟网友所指的估计是QCD（及其它一些渐近自由场论）中重整化质量参数在高能极限下趋于零这一结果（比如Weinberg的18.4节）。不过如sage兄所说，这与是否存在t介子并无关系。
宠辱不惊，看庭前花开花落
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sage 发表文章数: 359
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重整化质量参数在高能极限下趋于零这一结果
Such a conclusion actually have almost no physical effect whatsoever in the Standard Model. 1) due to chiral symmetry, the renormalization of the quark mass could only be logarithmic and proportional to itself. Therefore, it will be renormalized to zero only when energy goes to infinity. We will never reach that fix point, ever. 2) At energies higher than the quark masses, it is a good approximation to treat them as massless anyway, regardless how their masses are renormalized.
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一叶孤舟 发表文章数: 38
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谢谢二位老师！这几天过节到外边转了转，未及时过来，实在抱歉！我说的“顶介子”只是一种推测，繁劳二位老师费心了，我接受二位老师的解释。看来，我所看的书籍有些过时了，急需更新，可是，我不知道从哪里能够得到关于QCD的最新知识，只好再劳二位老师给以指点了。
忍一时风平浪静 退一步海阔天空
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landscape 发表文章数: 35
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能量趋于无穷时 的确要求其质量趋进于零 而ＱＣＤ耦合常数在顶夸克质量１７５ＧeV时 已经相当的小了 顶夸克的半衰期约为10^-24秒 呵呵 要么还没来得及就衰变掉 要么就无法耦合 你看看渐进自由理论吧
而且我也想请教sage和卢两位老师一个问题 u d s 夸克质量项可以破缺强相互作用的各种对称性 那么c b t呢？ 另外我想知道现在关于有效场论的评价如何？：） 很想弄明白D-T、D-D反应的截面（纯理论计算的，现在我们的数据是用唯象模型做的）。 你真美啊，请停一停！
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landscape 发表文章数: 35
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更正一下
我第一行说的质量是指重正化的那个质量参数m(u)->0,当 u->无穷时。 只有这样u很大时 才可以忽略其质量。 你真美啊，请停一停！
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sage 发表文章数: 359
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u d s 夸克质量项可以破缺强相互作用的各种对称性, 那么c b t呢？
it depends on which symmetry you are referring to as the symmetry of QCD. If you are talking about SU(3) color, then quark masses do not break color. I think you are probably talking about the SU(3)_L X SU(3)_R flavor symmetry, by (u,d,s). The reason we are talking about such a symmetry is that it is indeed an approximate symmetry. The masses of u, d, s break it, but only by a small amount since their masses are smaller than the QCD scale. In this sense, it is useful to talk about this approximate symmetry. On the other hand, masses of c, b, t are all bigger than the QCD scale. Therefore, it is not useful to think there is a symmtry since it is broken maximally. In fact, in the case of b, it is useful to think of another expansion, called heavy quark expansion, which treat QCD scale as an perturbation. 另外我想知道现在关于有效场论的评价如何？：） It is great. It is the correct language to think about field theory. IT is very useful, such as the argument I gave above. 很想弄明白D-T、D-D反应的截面（纯理论计算的，现在我们的数据是用唯象模型做的）。
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landscape 发表文章数: 35
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谢谢sage的答复
我的确主要是指SU（3）_L*SU(3)_R近似对称 当时我想的是既然能用u d s构架SU(3)，那么能不能在更高的能标下用上所有的夸克来构造更严格的对称？ 我对氘—氘、氘—氚反应的低能区（1个GeV以下）的截面以及其他数据感兴趣 精度要求也相当的高：） 目前我还不知道用什么方法去算它比较好 NLO？QEFT？ 老板又要k我了 先撤 再次叩首 你真美啊，请停一停！
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sage 发表文章数: 359
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我对氘—氘、氘—氚反应的低能区（1个GeV以下）的截面以及其他数据感兴趣
精度要求也相当的高：） I don't think you can get very far in this case by doing direct calculations. The energy involved are close to the masses and QCD scale. Therefore, there is probably no small parameter you could expand with.
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walk_f 发表文章数: 31
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补充一点(个人理解):
感觉短寿命并不一定导致不能组成束缚态 比如 自由中子短寿命 但可以和质子组成束缚态 并且 中子的短寿命也是因为弱衰变 组成束缚则倚赖强作用 这点似乎说明 本身短寿命比可能组成的束缚态的低不能否定能组成束缚态 这点不知道怎么解释? (介子只有2个夸克 可能比较好下结论 数目多起来的时候 可能比较难) "我对氘—氘、氘—氚反应的低能区（1个GeV以下）的截面以及其他数据感兴趣 精度要求也相当的高：）" -- 楼主是怎么算这个散射截面的? 是用类似量子力学中散射的处理方法么?(假定唯象势 弹性散射) 还是在夸克层次上(或者是作伪pn束缚态)做计算? very thing will be OK
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landscape 发表文章数: 35
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I don't think you can get very far in this case by doing direct calculations. The energy involved are close to the masses and QCD scale.
Thank sage for answers very much.But it is our task.we need more accurate data for some spacial users:-),such as double-differential cross section,energy spectrum and angular distributions,etc. Therefore, there is probably no small parameter you could expand with. This is my major barrier.i have no ideas:(. i don't know whether have non-perturbative theory to calculate this realistic problem?:) THANK YOU VERY MUCH FOR ADVICES. BEST WISHES FOR YOU. landscape 你真美啊，请停一停！
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sage 发表文章数: 359
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比如 自由中子短寿命 但可以和质子组成束缚态
? First, the question is whether something forms a bound state at all (stable or not). It is a question of comparing time (or energy scales). life time of free neutron is about 10^3 sec. binding energy of a nuclear bound state is about MeV, which corresponds to a time scale of 10^-20 sec. Therefore, decaying of neutron (if it happens at all), won't be even remotely close to affect the formation of a boundstate. Neutron will decide whether it should form a bound state way way before it realizes that it would decay. Second, there is the question of whether a bound state is stable. Of course, once you form a bound states, since the binding energy is about the same order as the free neutron and free proton mass difference, it is not surprising at all that neutron is not allowed to decay at all. This is VERY VERY different from the case of top quark. It's life time is shorter than the typical binding energy of a bound state or strong QCD interactions. It decays without even aware of there is a possibility of forming a bound state. I don't see why the neutron example is relevant for "top meson" at all. 并且 中子的短寿命也是因为弱衰变 组成束缚则倚赖强作用 这点似乎说明 本身短寿命比可能组成的束缚态的低不能否定能组成束缚态 这点不知道怎么解释? (介子只有2个夸克 可能比较好下结论 数目多起来的时候 可能比较难)
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walk_f 发表文章数: 31
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"First, the question is whether something forms a bound state at all (stable or not). It is a question of comparing time (or energy scales).
life time of free neutron is about 10^3 sec. binding energy of a nuclear bound state is about MeV, which corresponds to a time scale of 10^-20 sec." -- en 为说明方便 标记一下 假设 A+B->(束缚成) C 这里有3个能标 A的衰变宽度E1(寿命),AB形成C的束缚能E2,C的衰变宽度(束缚态的稳定性)E3, sage上面说的不能形成束缚态的原因是E1>>E2,我理解是这个意思?(不知道对否) 恩 这里我可能原来理解错了 把"The time scale it takes for a quark to form a meson is about the inverse of the QCD scale 1/(several hundred MeV)." 这个理解成E3了.所以举了中子和质子形成D的情况(E2>E1>E3的情况) 从量子力学的情况来看 能否形成束缚态好像和束缚能的具体大小(E2)无关 而只和作用形式有关(不知道我这个理解对不) 如果是这样的话 条件E1>>E2感觉难以用量子力学的图像来对应理解 (嗯 量子力学图像本身用在短寿命粒子上面可能不太合适 但是对束缚态 离开量子力学的图像 又感觉挺难直观理解) "Second, there is the question of whether a bound state is stable." -- 我的感觉是E3是很难由E1 和E2 来反应出来 感觉基本上应该无关 (不知道这里有没有联系?) "This is VERY VERY different from the case of top quark. It's life time is shorter than the typical binding energy of a bound state or strong QCD interactions. It decays without even aware of there is a possibility of forming a bound state." -- "or strong QCD interactions" 这指的是束缚态介子的衰变宽度么 还是就是束缚能? very thing will be OK
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walk_f 发表文章数: 31
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嗯 补充一下
我觉得 top夸克不能形成介子 为什么不直接用其相应的E3很大来解释呢? 形成的介子本身寿命很短很短(和其他轻夸克介子完全不一样) 而撇开夸克本身的衰变寿命 (虽然两者可能有点关) - 一点疑惑 very thing will be OK
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sage 发表文章数: 359
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恩 这里我可能原来理解错了 把"The time scale it takes for a quark to form a meson is about the inverse of the QCD scale 1/(several hundred MeV)." 这个理解成E3了.所以举了中子和质子形成D的情况(E2>E1>E3的情况)
从量子力学的情况来看 能否形成束缚态好像和束缚能的具体大小(E2)无关 而只和作用形式有关(不知道我这个理解对不) 如果是这样的话条件E1>>E2感觉难以用量子力学的图像来对应理解 (嗯 量子力学图像本身用在短寿命粒子上面可能不太合适 但是对束缚态离开量子力学的图像 又感觉挺难直观理解) "Second, there is the question of whether a bound state is stable." -- 我的感觉是E3是很难由E1 和E2 来反应出来 感觉基本上应该无关 (不知道这里有没有联系?) "This is VERY VERY different from the case of top quark. It's life time is shorter than the typical binding energy of a bound state or strong QCD interactions. It decays without even aware of there is a possibility of forming a bound state." -- "or strong QCD interactions" 这指的是束缚态介子的衰变宽度么 还是就是束缚能? ============================================================================== The energy scale associated with a particular process determines, according to quantum mechanics, the time scale of such a process. For top quark, E1>>E2, therefore, top decays much earlier than it even have chance to form bound state with anything. E3 is not relevant in this discussion.
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walk_f 发表文章数: 31
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"For top quark, E1>>E2, therefore, top decays much earlier
than it even have chance to form bound state with anything. E3 is not relevant in this discussion." - 这点也是我困惑的 因为在量子力学的图像中 E1=0; 而束缚能则是可以很大 或者趋于0 嗯 我是感觉量子力学里面 能否形成束缚态基本上与束缚能大小是无关 主要是和作用形式有关 所以 在讨论这些本身就会衰变的粒子能否形成束缚态时候 要是用E1>>E2(不能形成)这个条件 感觉回不到量子力学的情况 而解释为E1<<E3(不能形成)这个条件 好像两者就一致了 very thing will be OK
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sage 发表文章数: 359
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- 这点也是我困惑的
因为在量子力学的图像中 E1=0 Why? Top quark will decay, therefore its width is not zero. ; 而束缚能则是可以很大 或者趋于0 嗯 我是感觉量子力学里面 能否形成束缚态基本上与束缚能大小是无关 主要是和作用形式有关 所以 在讨论这些本身就会衰变的粒子能否形成束缚态时候 要是用E1>>E2(不能形成)这个条件 感觉回不到量子力学的情况 而解释为E1<<E3(不能形成)这个条件 好像两者就一致了 For two STABLE particles, they will form a bound state as long as binding energy is less than zero (or the total energy of the boundstate is less than two separate free particles). Size of binding energy does not control whether a bound state will form. However, it does control how long it takes to form a bound state. For example, it controls, starting from two particles with random momentum, how long it will take to form a spatially localized state... For mesonic states, the typical time scale of forming a bound state is always not very far away from the QCD scale. However, I will have to say it again, top quark decays. and, it decays much faster than the time it takes for it to form any meaningful bound state. Therefore, it is will not form any meaningful or well defined bound state.
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walk_f 发表文章数: 31
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"Why? Top quark will decay, therefore its width is not zero."
-嗯 E1指量子力学情况下的E1 "For two STABLE particles, they will form a bound state as long as binding energy is less than zero (or the total energy of the boundstate is less than two separate free particles)." -- 嗯 我的问题是 束缚能<0是形成束缚态的必要条件么? 我是对这个有点困惑:( 量子力学忘记了:( 嗯 感觉挺难证明这个 另外激发态的存在E_all>m_1+m_2也是和这个有点不一样? 不知道binding energy是否专门针对基态 嗯 我的想法是想先弄清除量子力学的情况下(E1=0的情况) 束缚能<0是不是形成束缚态的必要条件?(量子力学忘光了 呵呵 偷懒的不想去查查资料 直接这里询问了) 然后再来看E1很大的情况下(比如 top夸克) 这个条件是否还是这样? 嗯 不过我实在是看不出来用E3<<E1这个条件做能形成束缚态的必要条件 有什么不对头的地方:( (感觉这个条件很平庸 也刚好可用 ) very thing will be OK
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sage 发表文章数: 359
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E1指量子力学情况下的E1
why is E1=0 in quantum mechanics?
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walk_f 发表文章数: 31
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"why is E1=0 in quantum mechanics?"
-- 量子力学处理的A+B形成束缚态C 默认A,B是稳定粒子 不衰变 对应的衰变宽度E1=0; 用量子力学来处理不稳定的粒子形成束缚态 应该不太合适 我的理解是 场论下面本身讨论束缚态太困难 只好拿量子力学这种特殊情况(E1=0)来类推top夸克了 very thing will be OK
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一叶孤舟 发表文章数: 38
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谢谢上边几位老师的讨论与教诲，这使我受益匪浅，请继续。
我得赶快再啃啃量子力学。 忍一时风平浪静 退一步海阔天空
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sage 发表文章数: 359
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-- 量子力学处理的A+B形成束缚态C 默认A,B是稳定粒子
Not necessarily. one can always incorperate the decay by introducing an imaginary part to Hamiltonian. 不衰变对应的衰变宽度E1=0; not a correct description to the top case. 用量子力学来处理不稳定的粒子形成束缚态 应该不太合适 nothing wrong with it. By the way, quantum field theory and quantum mechanics are not independent physics. They are all just quantum mechanics. IT is just more convenient to use field theory language when the number of degrees of freedom become large. Either language you use, the conclusion should be the same. Anyway, the bottom line of the question is top-quark decays before it forms a bound state with something else. That's it. it is a physical result independent of whichever language one uses.
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walk_f 发表文章数: 31
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Not necessarily. one can always incorperate the decay by introducing an imaginary part to Hamiltonian.
-- 这个尔米性有些问题吧 very thing will be OK
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sage 发表文章数: 359
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-- 这个尔米性有些问题吧
Hamiltonian will certainly have to be hermitian, on the FULL hilbert space. On the other hand, it certainty does not have to be hermitian (and it is not hermitian) in the sub-Hilbert space spanned by some particle which can decay into others.
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walk_f 发表文章数: 31
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"Hamiltonian will certainly have to be hermitian, on the FULL hilbert space. On the other hand, it certainty does not have to be hermitian (and it is not hermitian) in the sub-Hilbert space spanned by some particle which can decay into others."
--- 这个是没问题 但是 现在的量子力学里面 有处理由不稳定粒子组成的束缚态情况么? 我好像没见过 :( 一直感觉没 :) 嗯 也许 H=H_01+H_02+H_int H 尔米 H_01 分别不尔米 倒是可能实现? 不过 没看到过有讨论这个的资料 也许 sage可以提供一些 : ) 怎么在场论下面理解束缚态问题 一直是让我困惑的问题 我觉得 和量子力学下面的图像比较起来 不是单纯自由度的不同的 恩 也许 sage可以说的更多些 呵呵 very thing will be OK
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